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(B) When the ball is released,the forces acting on it are the buoyant force $F_B$ (upwards),the weight $mg$ (downwards),and the viscous drag force $F_

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with a decreasing acceleration

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(B) When the ball is released,the forces acting on it are the buoyant force $F_B$ (upwards),the weight $mg$ (downwards),and the viscous drag force $F_v$ (downwards,as the ball moves up).$F_B = V(2\rho)g = 2V\rho g$$mg = V\rho g$Net upward force $F_{net} = F_B - mg - F_v = 2V\rho g - V\rho g - F_v = V\rho g - F_v$.As the ball moves up,its velocity $v$ increases,which causes the viscous force $F_v$ (where $F_v \propto v$) to increase.Since $F_{net} = V\rho g - F_v$,as $F_v$ increases,the net force $F_{net}$ decreases.According to Newton's second law,$a = F_{net} / m$. As $F_{net}$ decreases,the acceleration $a$ of the ball decreases.

$A$ ball of density $\rho$ is released from deep inside a viscous liquid of density $2\rho$. It will move up:

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